What is the Ito equation of $ \ddot{x}(t)+\kappa x(t)=h \dot{B}(t) + \ddot{B}(t) $?

Question

Let $B(t)$ be a 1-dimensional Brownian motion. Consider the equation $$ \ddot{x}(t)+\kappa x(t)=h \dot{B}(t) \quad \text { on } t \geq 0 $$ where $\kappa$ and $h$ are positive constants. By introducing the new variable $y(t)=\dot{x}(t)$, the corresponding Itô equation is $$ d\left[\begin{array}{l} x(t) \\ y(t) \end{array}\right]=A\left[\begin{array}{l} x(t) \\ y(t) \end{array}\right] d t+\left[\begin{array}{l} 0 \\ h \end{array}\right] d B(t) $$ where $$ A=\left[\begin{array}{cc} 0 & 1 \\ -\kappa & 0 \end{array}\right] $$ But what if we add the second derivative of Brownian motion to the equation? I mean, e.g.: $$ \ddot{x}(t)+\kappa x(t)=h \dot{B}(t) + \ddot{B}(t) \quad \tag{*} $$ What is the corresponding Ito equation of (*)?

Answer 1

One variant: To reduce the order in the equation, combine the derivative terms. The substitution will then also have lower order derivatives. $$ \frac{d}{dt}(\dot x-hB-\dot B)+x=0 $$ Thus set $\dot x=v+hB+\dot B$ to get $\dot v+κx=0$ or more well-defined in shortened Ito integral form $$ dx = (v+hB)\,dt+dB \\ dv=-κx\,dt. $$ One could shift some terms around to get other first-order systems, here mainly reducing the substitution to $\dot x=v+\dot B$ so that $\dot v+κx=h\dot B$ or $$ dx=v\,dt+dB\\dv = -κx\,dt+h\,dB $$